Force, Find the Angle Given 2 Forces of Equal Magnitude, but Different Direction

by Simon777
Tags: angle, direction, equal, force, forces, magnitude
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 P: 35 1. The problem statement, all variables and given/known data Two horses pull horizontally on ropes attached to a tree stump. Each horse pulls with a force of magnitude F. If the resultant force (vector R) has the magnitude R = 1.51 F, what is the angle (in degrees) between the two ropes? 2. Relevant equations 3. The attempt at a solution Ry= 0+cos Rx= 1+ sin R= 1.51 sin=cos/1.54 tan= 1/1.54 angle= 33 degrees 33 * 2 = total angle for both of 66 degrees
 P: 35 I'm on my 5th hour for this problem and still can't get it. I realize it's necessary to assign an arbitrary value of say 1 to F. That gives F=1 and R=1.51, but everything I've tried to do with that information has failed.
 P: 58 Using vector addition... We don't know whether the three vectors make a right triangle (when putting the two F's nose to tail), so it's easier to divide them into 2 triangles. We'll call the new segment G. So now F becomes the Hypotenuse, G is Opposite, and 1.51/2 F is the adjacent (divided in half because it's now the adjacent on 2 triangles instead of just 1). F sin(angle) = 1.51/2 F => divide out the F => arcsin(1.51/2) = angle = 49.025 degrees That's just half of your angle though since you split it into two triangles 49.025*2 = 98.05 degrees
P: 35
Force, Find the Angle Given 2 Forces of Equal Magnitude, but Different Direction

 Quote by Allenman Using vector addition... We don't know whether the three vectors make a right triangle (when putting the two F's nose to tail), so it's easier to divide them into 2 triangles. We'll call the new segment G. So now F becomes the Hypotenuse, G is Opposite, and 1.51/2 F is the adjacent (divided in half because it's now the adjacent on 2 triangles instead of just 1). F sin(angle) = 1.51/2 F => divide out the F => arcsin(1.51/2) = angle = 49.025 degrees That's just half of your angle though since you split it into two triangles 49.025*2 = 98.05 degrees
The professor gave an answer of 81.9 degrees.
 P: 58 Hmmmm maybe someone else will chime in... I don't see anything wrong with what I did, maybe I'm overlooking something.
P: 35
 Quote by Allenman Hmmmm maybe someone else will chime in... I don't see anything wrong with what I did, maybe I'm overlooking something.
I really appreciate the response, I initially did what you did and was told to rethink it and that it was 81.9. That's what's been making me go insane. I don't know why this method is wrong.
P: 35
 Quote by Allenman Hmmmm maybe someone else will chime in... I don't see anything wrong with what I did, maybe I'm overlooking something.
You had it right and explained it right, but used arcsin instead of arccos. I get how to do this now, thank you so much for helping me finally understand this.

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